b
When the number of hybrid orbitals, \(H\) is \(4\), the hybridization is \(s p^{3}\) .

\(H=\frac{1}{2}[V+M-C+A]\)

where, \(V=\) number of valence electrons of central atom

\(M=\) number of monovalent atoms

\(C=\) total positive charge

\(\mathrm{A}=\) negative charge

\(H=\frac{1}{2}[4+4-0+0]=4\)

thus \(s p^{3}\) hybridization

For \(S F_{4}\) \(H=\frac{1}{2}[6+4-0+0]=5\)

thus \(s p^{3} d\) hybridization

For \(B F_{4}\)

\(H=\frac{1}{2}[3+4-0+1]=4\)

thuss \(p^{3}\) hybridization

For \(N H_{4}^{+}\) \(H=\frac{1}{2}[5+4-1+0]=4\)

thuss \(s p^{3}\) hybridization Thus, only in \(S F_{4}\)

the central atom does not have \(sp ^3\) hybridization.